//Dynamic Programming solution for the Egg Dropping Puzzle
public class EggDropping
{
	
	// min trials with n eggs and m floors 
	
	private static int minTrials(int n, int m)
	{
	    
		int eggFloor[][] = new int[n+1][m+1];
		int result, x;
		
		for (int i = 1; i <= n; i++)
		{
		    eggFloor[i][0] = 0;   // Zero trial for zero floor.
		    eggFloor[i][1] = 1;   // One trial for one floor 
		}
		
	        // j trials for only 1 egg
		
                for (int j = 1; j <= m; j++)
			 eggFloor[1][j] = j;
		
		// Using bottom-up approach in DP
		
   		 for (int i = 2; i <= n; i++)
		  {
			for (int j = 2; j <= m; j++)
			{
				eggFloor[i][j] = Integer.MAX_VALUE;
				for (x = 1; x <= j; x++)
				{
					result = 1 + Math.max(eggFloor[i-1][x-1], eggFloor[i][j-x]);
          
					//choose min of all values for particular x
        				 if (result < eggFloor[i][j])
					     eggFloor[i][j] = result;
				}
			}
		}
		  
		return eggFloor[n][m];
	}
		
	//testing program
	public static void main(String args[])
	{
	      int n = 2, m = 4;
              //result outputs min no. of trials in worst case for n eggs and m floors
              int result = minTrials(n, m);
	      System.out.println(result); 
	}
}
